Assume $f$ is continuous on $[a,b]$, derivable on $(a,b)$. $f(b)gt f(a)$

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Assume $f$ is continuous on $[a,b]$, derivable on $(a,b)$. $f(b)gt f(a)$, $c=fracf(b)-f(a)b-a$. Then there must hold one of the claims:

1.$forall xin[a,b] $, $f(x)-f(a)=c(x-a)$

2.$exists xi in (a,b), f'(xi)gt c$.

---

Well I think the way is that if we deny one then another must hold.

So I choose to deny 2.

Now if $forall x,f'(x)le c$ , we will claim it must hold $f'(x)=c$. Suppose there is a $eta$ such that $f'(eta) lt c$. Denote $F(x)=f(x)-f(a)-fracf(b)-f(a)b-a(x-a)$. So $F'(x)=f'(x)-fracf(b)-f(a)b-ale 0$. But $F(a)=F(b)=0$ and $F(x)neq0$.Hence $f'(x)=c$ and it follows 1. holds.

Now was I make it strictly. I'm not sure I had give a strict proof.

calculus

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asked Jul 21 at 15:21

Jaqen Chou

1416

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up vote
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favorite

1

Assume $f$ is continuous on $[a,b]$, derivable on $(a,b)$. $f(b)gt f(a)$, $c=fracf(b)-f(a)b-a$. Then there must hold one of the claims:

1.$forall xin[a,b] $, $f(x)-f(a)=c(x-a)$

2.$exists xi in (a,b), f'(xi)gt c$.

---

Well I think the way is that if we deny one then another must hold.

So I choose to deny 2.

Now if $forall x,f'(x)le c$ , we will claim it must hold $f'(x)=c$. Suppose there is a $eta$ such that $f'(eta) lt c$. Denote $F(x)=f(x)-f(a)-fracf(b)-f(a)b-a(x-a)$. So $F'(x)=f'(x)-fracf(b)-f(a)b-ale 0$. But $F(a)=F(b)=0$ and $F(x)neq0$.Hence $f'(x)=c$ and it follows 1. holds.

Now was I make it strictly. I'm not sure I had give a strict proof.

calculus

share|cite|improve this question

asked Jul 21 at 15:21

Jaqen Chou

1416

add a comment | 

up vote
-1
down vote

favorite

1

up vote
-1
down vote

favorite

1

1

Assume $f$ is continuous on $[a,b]$, derivable on $(a,b)$. $f(b)gt f(a)$, $c=fracf(b)-f(a)b-a$. Then there must hold one of the claims:

1.$forall xin[a,b] $, $f(x)-f(a)=c(x-a)$

2.$exists xi in (a,b), f'(xi)gt c$.

---

Well I think the way is that if we deny one then another must hold.

So I choose to deny 2.

Now if $forall x,f'(x)le c$ , we will claim it must hold $f'(x)=c$. Suppose there is a $eta$ such that $f'(eta) lt c$. Denote $F(x)=f(x)-f(a)-fracf(b)-f(a)b-a(x-a)$. So $F'(x)=f'(x)-fracf(b)-f(a)b-ale 0$. But $F(a)=F(b)=0$ and $F(x)neq0$.Hence $f'(x)=c$ and it follows 1. holds.

Now was I make it strictly. I'm not sure I had give a strict proof.

calculus

share|cite|improve this question

asked Jul 21 at 15:21

Jaqen Chou

1416

Assume $f$ is continuous on $[a,b]$, derivable on $(a,b)$. $f(b)gt f(a)$, $c=fracf(b)-f(a)b-a$. Then there must hold one of the claims:

1.$forall xin[a,b] $, $f(x)-f(a)=c(x-a)$

2.$exists xi in (a,b), f'(xi)gt c$.

---

Well I think the way is that if we deny one then another must hold.

So I choose to deny 2.

Now if $forall x,f'(x)le c$ , we will claim it must hold $f'(x)=c$. Suppose there is a $eta$ such that $f'(eta) lt c$. Denote $F(x)=f(x)-f(a)-fracf(b)-f(a)b-a(x-a)$. So $F'(x)=f'(x)-fracf(b)-f(a)b-ale 0$. But $F(a)=F(b)=0$ and $F(x)neq0$.Hence $f'(x)=c$ and it follows 1. holds.

Now was I make it strictly. I'm not sure I had give a strict proof.

calculus

share|cite|improve this question

asked Jul 21 at 15:21

Jaqen Chou

1416

share|cite|improve this question

share|cite|improve this question

asked Jul 21 at 15:21

Jaqen Chou

1416

asked Jul 21 at 15:21

Jaqen Chou

1416

asked Jul 21 at 15:21

Jaqen Chou

1416

1416

add a comment | 

add a comment | 

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If the first statement holds, then we have $$f(x)=cx-ac+f(a),~~~forall x in [a,b].$$
Thus, $$f'(x)=c,~~~forall x in (a,b),$$ which contradicts the second statement. Therefore, the second statement does not hold.

Now, let's repeat the fact by logical language. Denote the first one as $A$, the sencond one as $B$. We have already showed that $A to barB.$ Thus, $B to barA.$In one word, $A$ and $B$ contradict each other.

But can you prove $barA to B$ ? Let's try it. Notice that $barA$ states that:

$$exists x_0 in [a,b],s.t.f(x_0)-f(a)neq c(x_0-a)$$

If $f(x_0)-f(a)> c(x_0-a),$ then by Lagrange mean value theorem, we have $$exists xi in (a,x_0), s.t. f'(xi)=fracf(x_0)-f(a)x_0-a>0,$$ which implies $B$.

If $f(x_0)-f(a)< c(x_0-a),$ then by Lagrange mean value theorem, we have $$exists xi in (x_0,b), s.t. f'(xi)=fracf(b)-f(x_0)b-x_0>fracf(b)-f(a)b-a=c,$$ which implies $B$.

We are done！

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edited Jul 21 at 17:28

answered Jul 21 at 16:20

mengdie1982

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If the first statement holds, then we have $$f(x)=cx-ac+f(a),~~~forall x in [a,b].$$
Thus, $$f'(x)=c,~~~forall x in (a,b),$$ which contradicts the second statement. Therefore, the second statement does not hold.

Now, let's repeat the fact by logical language. Denote the first one as $A$, the sencond one as $B$. We have already showed that $A to barB.$ Thus, $B to barA.$In one word, $A$ and $B$ contradict each other.

But can you prove $barA to B$ ? Let's try it. Notice that $barA$ states that:

$$exists x_0 in [a,b],s.t.f(x_0)-f(a)neq c(x_0-a)$$

If $f(x_0)-f(a)> c(x_0-a),$ then by Lagrange mean value theorem, we have $$exists xi in (a,x_0), s.t. f'(xi)=fracf(x_0)-f(a)x_0-a>0,$$ which implies $B$.

If $f(x_0)-f(a)< c(x_0-a),$ then by Lagrange mean value theorem, we have $$exists xi in (x_0,b), s.t. f'(xi)=fracf(b)-f(x_0)b-x_0>fracf(b)-f(a)b-a=c,$$ which implies $B$.

We are done！

share|cite|improve this answer

edited Jul 21 at 17:28

answered Jul 21 at 16:20

mengdie1982

2,912216

add a comment | 

up vote
0
down vote

If the first statement holds, then we have $$f(x)=cx-ac+f(a),~~~forall x in [a,b].$$
Thus, $$f'(x)=c,~~~forall x in (a,b),$$ which contradicts the second statement. Therefore, the second statement does not hold.

Now, let's repeat the fact by logical language. Denote the first one as $A$, the sencond one as $B$. We have already showed that $A to barB.$ Thus, $B to barA.$In one word, $A$ and $B$ contradict each other.

But can you prove $barA to B$ ? Let's try it. Notice that $barA$ states that:

$$exists x_0 in [a,b],s.t.f(x_0)-f(a)neq c(x_0-a)$$

If $f(x_0)-f(a)> c(x_0-a),$ then by Lagrange mean value theorem, we have $$exists xi in (a,x_0), s.t. f'(xi)=fracf(x_0)-f(a)x_0-a>0,$$ which implies $B$.

If $f(x_0)-f(a)< c(x_0-a),$ then by Lagrange mean value theorem, we have $$exists xi in (x_0,b), s.t. f'(xi)=fracf(b)-f(x_0)b-x_0>fracf(b)-f(a)b-a=c,$$ which implies $B$.

We are done！

share|cite|improve this answer

edited Jul 21 at 17:28

answered Jul 21 at 16:20

mengdie1982

2,912216

add a comment | 

up vote
0
down vote

up vote
0
down vote

If the first statement holds, then we have $$f(x)=cx-ac+f(a),~~~forall x in [a,b].$$
Thus, $$f'(x)=c,~~~forall x in (a,b),$$ which contradicts the second statement. Therefore, the second statement does not hold.

Now, let's repeat the fact by logical language. Denote the first one as $A$, the sencond one as $B$. We have already showed that $A to barB.$ Thus, $B to barA.$In one word, $A$ and $B$ contradict each other.

But can you prove $barA to B$ ? Let's try it. Notice that $barA$ states that:

$$exists x_0 in [a,b],s.t.f(x_0)-f(a)neq c(x_0-a)$$

If $f(x_0)-f(a)> c(x_0-a),$ then by Lagrange mean value theorem, we have $$exists xi in (a,x_0), s.t. f'(xi)=fracf(x_0)-f(a)x_0-a>0,$$ which implies $B$.

If $f(x_0)-f(a)< c(x_0-a),$ then by Lagrange mean value theorem, we have $$exists xi in (x_0,b), s.t. f'(xi)=fracf(b)-f(x_0)b-x_0>fracf(b)-f(a)b-a=c,$$ which implies $B$.

We are done！

share|cite|improve this answer

edited Jul 21 at 17:28

answered Jul 21 at 16:20

mengdie1982

2,912216

If the first statement holds, then we have $$f(x)=cx-ac+f(a),~~~forall x in [a,b].$$
Thus, $$f'(x)=c,~~~forall x in (a,b),$$ which contradicts the second statement. Therefore, the second statement does not hold.

Now, let's repeat the fact by logical language. Denote the first one as $A$, the sencond one as $B$. We have already showed that $A to barB.$ Thus, $B to barA.$In one word, $A$ and $B$ contradict each other.

But can you prove $barA to B$ ? Let's try it. Notice that $barA$ states that:

$$exists x_0 in [a,b],s.t.f(x_0)-f(a)neq c(x_0-a)$$

If $f(x_0)-f(a)> c(x_0-a),$ then by Lagrange mean value theorem, we have $$exists xi in (a,x_0), s.t. f'(xi)=fracf(x_0)-f(a)x_0-a>0,$$ which implies $B$.

If $f(x_0)-f(a)< c(x_0-a),$ then by Lagrange mean value theorem, we have $$exists xi in (x_0,b), s.t. f'(xi)=fracf(b)-f(x_0)b-x_0>fracf(b)-f(a)b-a=c,$$ which implies $B$.

We are done！

share|cite|improve this answer

edited Jul 21 at 17:28

answered Jul 21 at 16:20

mengdie1982

2,912216

share|cite|improve this answer

share|cite|improve this answer

edited Jul 21 at 17:28

edited Jul 21 at 17:28

edited Jul 21 at 17:28

answered Jul 21 at 16:20

mengdie1982

2,912216

answered Jul 21 at 16:20

mengdie1982

2,912216

answered Jul 21 at 16:20

mengdie1982

2,912216

2,912216

add a comment | 

add a comment | 

 

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